Page 80 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Hamiltonian formulation 67

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Example 2.14 Consider the simplest case where f (x, u, ξ)= f (ξ) with f > 0
(or more generally f is strictly convex) and lim |ξ|→∞ f (ξ) /ξ =+∞. The Euler-
Lagrange equation and its integrated form are

d
0
0
0
0
[f (u )] = 0 ⇒ f (u )= λ = constant.
dx
The Hamiltonian is given by
∗
H (v)= f (v)= sup {vξ − f (ξ)} .
ξ
The associated Hamiltonian system is
⎧
0
⎨ u = f (v)
∗0
⎩ 0
v =0 .
0
We find trivially that (λ and µ denoting some constants) v = λ and hence
(compare with Case 2.3)
∗0
u (x)= f (λ) x + µ.
Example 2.15 We now look for the slightly more involved case where f (x, u, ξ)=
f (x, ξ) with the appropriate hypotheses. The Euler-Lagrange equation and its
integrated form are
d
0
[f ξ (x, u )] = 0 ⇒ f ξ (x, u )= λ = constant.
0
dx
The Hamiltonian of f, is given by
H (x, v)= sup {vξ − f (x, ξ)} .
ξ
The associated Hamiltonian system is
⎧
⎨ u (x)= H v (x, v (x))
0
v =0 .
⎩
0
The solution is then given by v = λ = constant and u (x)= H v (x, λ).
0
Example 2.16 We consider next the more difficult case where f (x, u, ξ)=
f (u, ξ) with the hypotheses of the theorem. The Euler-Lagrange equation and
its integrated form are
d
0
0
0
0
[f ξ (u, u )] = f u (u, u ) ⇒ f (u, u ) − u f ξ (x, u )= λ = constant.
0
dx